Branch-Circuit, Feeder and Service Calculations, Part XXV

by Charles R. Miller
Published: March 2008

Article 220 – Load Calculations

220.55 Electric Ranges and Other Cooking Appliances—Dwelling Unit

Understanding how to perform load calculations in
accordance with the National Electrical Code (NEC) is essential for the
professional electrician. Loads must be calculated before installing
branch-circuits, feeders and services. Chapter 2 in the NEC contains 10
articles. Article 220 contains requirements for calculating
branch-circuit, feeder and service loads. Results from calculations in
Article 220 are used with requirements from other articles to find
conductor sizes and ampere ratings for overcurrent protective devices.
For example, results from calculations in Parts III, IV and V of
Article 220 are used with the provision in 215.2(A)(1) to find the
minimum feeder-circuit conductor size. Feeder conductors shall have an
ampacity not less than required to supply the load as calculated in
Parts III, IV and V of Article 220 [215.2(A)(1)]. This same section
continues by specifying that the minimum feeder-circuit conductor size,
before the application of any adjustment or correction factors, shall
have an allowable ampacity not less than the noncontinuous load plus
125 percent of the continuous load. Likewise, results (from
calculations in Article 220) are used with provisions in 215.3 to find
the minimum size fuse or breaker permitted for feeders.

Last
month’s Code in Focus covered electric cooking equipment in 220.55.
This month, the discussion continues with calculating loads for
electric ranges and other cooking appliances in dwelling units.

One
example last month calculated the demand for 25 5-kW counter-mounted
cooking units (cooktops). The demand factor for those 5-kW cooktops was
found in Column B, which is for household cooking appliances rated at
least 3˝ kW but not more than 8ľ kW. The demand factor percent then was
multiplied by the total kilowatts of the cooktops to give the service
demand load. Appliances can have different ratings but still be within
the limits of one column. For example, what is the service demand load
for five 3˝-kW wall-mounted ovens, five 5-kW counter-mounted cooking
units and 10 8-kW ranges? Because none of the appliances are rated less
than 3˝ kW or more than 8ľ kW, the demand factor will come from Column
B. The first step is to find the total number of units (5 + 5 + 10 =
20). Now, find the demand factor percent across from 20 units (28
percent). Next, find the total kilowatts of all the appliances. The
ovens have a total rating of 17˝ kW (5 × 3.5 = 17.5). The cooktops have
total rating of 25 kW (5 × 5 = 25). The ranges have a total rating of
80 kW (10 × 8 = 80). The combined rating of all the appliances is 122.5
kW (17.5 + 25 + 80 = 122.5). Finally, multiply the total kilowatt load
by the demand factor percent (122.5 × 28% = 34.3). The service demand
load for five 3˝-kW wall-mounted ovens, five 5-kW -counter-mounted
cooking units and 10 8-kW ranges is 34.3 kW (see Figure 1).

As discussed last month, any time either Column A or B
can be used to calculate household cooking appliances, Column C can be
used. This is because Column C is for household cooking appliances “not
over 12 kW rating.” After calculating the cooking equipment load in
either Column A or B, compare the load with the demand load in Column
C, and then select the lower of the two loads. For example, what is the
service demand load for five 3˝-kW wall-mounted ovens, five 5-kW
counter-mounted cooking units and 15 8-kW ranges? Because of their
ratings, the demand factor percent will be from Column B. The total
number of units is 25 (5 + 5 + 15 = 25). The demand factor percent
across from 25 units is 26 percent. The ovens have a total rating of
17.5 kW (5 × 3.5 = 17.5). The cooktops have a total rating of 25 kW (5
× 5 = 25). The ranges have a total rating of 120 kW (15 × 8 = 120). The
combined rating of all the appliances is 162˝ kW (17.5 + 25 + 120 =
162.5). Now, multiply the total kilowatt load by the demand factor
percent for 25 units (162.5 × 26% = 42.25). The demand load from Column
B is 42.25 kW. Next, find the demand load in Column C for 25 units (40
kW). Finally, compare the loads from both columns, and select the
lower. With this example, the lower number is from Column C. The
service demand load for five 3˝-kW wall-mounted ovens, five 5-kW
counter-mounted cooking units and 15 8-kW ranges is 40 kW (see Figure
2).

Look back at the first example, and compare the
calculated load from Column B with the maximum demand from Column C.
The total number of units from the first example was 20. The demand
load from Column C for 20 units is 35 kW. The calculated load from
Column B for five 3˝-kW wall-mounted ovens, five 5-kW counter-mounted
cooking units and 10 8-kW ranges was 34.3 kW. In the first example, the
lower of the two demand loads was from Column B (see Figure 3).

A number of electricians and contractors will find
this series on branch-circuit, feeder and service calculations very
helpful when preparing for a journeyman or master electrician’s
examination. One type of exam question pertaining to household cooking
equipment that often is misunderstood and, therefore, answered
incorrectly includes or uses the word minimum. This type of question
can be tricky because of the wording. Below is an example of this type
of question.

What is the minimum demand load in kilowatts for 10 8-kW ranges?

A. 20 kW B. 25 kW C. 27.2 kW D. 80 kW

The
ranges have a total rating of 80 kW (10 × 8 = 80). The demand factor
percent across from 10 units in Column B is 34 percent. The calculated
load from Column B is 27.2 kW (80 × 34% = 27.2). The demand load in
Column C for 10 units is 25 kW. Because the word minimum is in this
question, it looks like the correct answer should come from Column B,
especially since the word maximum is in the heading of Column C.
Because it can lead to confusion, disregard the word minimum in this
question. In accordance with Table 220.55, when comparing Columns B and
C, it is permissible to select the lower of the two. The demand load
for 10 8-kW ranges is 25 kW. The correct answer is B (see Figure 4).

Another type of exam question pertaining to household
cooking equipment that often is answered incorrectly includes or uses
the word maximum. Below is an example of this type of question.

What is the maximum demand load in kilowatts for five 8-kW ranges?

A. 18 kW B. 20 kW C. 25 kW D. 40 kW

The
ranges have a total rating of 40 kW (5 × 8 = 40). The demand factor
percent across from five units in Column B is 45 percent. The
calculated load from Column B is 18 kW (40 × 45% = 18). The demand load
in Column C for five units is 20 kW. With this question, it looks like
the correct answer must come from Column C because the question is
asking for the maximum demand. The maximum demand load required by
Table 220.55 is the lower of the two columns. Disregard the word
maximum in this question. The demand load for five 8-kW ranges is 18
kW. The correct answer is A (see Figure 5).

Next month’s column continues the discussion of feeder and service load calculations.

MILLER, owner of Lighthouse Educational
Services, teaches classes and seminars on the electrical industry. He
is the author of “Illustrated Guide to the National Electrical Code”
and NFPA’s “Electrical Reference.” He can be reached at 615.333.3336, charles@charlesRmiller.com or www.charlesRmiller.com.